I'm trying to determine the Thevenin equivalent voltage and resistance across this simple circuit below. I understand the theory but can get difficult when trying to apply it.

Voltage divider circuit and its thevenin equivalance

Using Kirchoff Loop and junction laws I came up with the following equation:

1) \$I1=I2+I3\$

2) \$Vin-I1R1-I2R2=0\$

3) \$Vin-I1R1-I3R3-Vth=0\$

4) \$I2R2=I3R3\$

Which I found the Thevenin voltage to be \$Vth=VinR3/(R3+R1)\$ and the resistance to be \$Rth = R1+(R2R3/R2+R3)\$

The resistance seems plausible but the voltage doesn't as I would expect R2 to be in the equation somewhere. Any help is appreciated as I am not sure what I have done wrong. Feel free to edit my question for clarity.


1 Answer 1


You are over complicating a simple problem. The Thevenin equivalent voltage is calculated with A and B open, so there is no current over R3. The Thevenin equivalent resistance is done making all the voltages passive.


Vth = Vin x R2 / (R1 + R2)

Rth = R1 // R2 + R3 (For Rth, R1 and R2 are in parallel)


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